Question

The number of integral values of \(n\), for which the equation \[ 3\cos x + 5\sin x = 2n + 1 \] has a solution, is:

• A. \(4\)
• B. \(6\)
• C. \(8\)
• D. \(10\)

Hint:

For equations of the form \(a\cos x + b\sin x = k\), always compare \(k\) with \(\pm\sqrt{a^2+b^2}\) to determine solvability.

Answer 1

The Correct Option is B

Concept:
An expression of the form \(a\cos x + b\sin x\) can be written as \[ \sqrt{a^2+b^2}\sin(x+\phi) \]
Hence, its range is \[ -\sqrt{a^2+b^2} \le a\cos x + b\sin x \le \sqrt{a^2+b^2}. \]
Step 1: Find the range of the left-hand side \[ 3\cos x + 5\sin x \] Its maximum value is: \[ \sqrt{3^2+5^2}=\sqrt{34} \] Hence, \[ -\sqrt{34} \le 3\cos x + 5\sin x \le \sqrt{34} \]
Step 2: Apply the condition for solution For the equation to have a solution, \[ -\sqrt{34} \le 2n+1 \le \sqrt{34} \]
Step 3: Solve the inequality \[ -\sqrt{34}-1 \le 2n \le \sqrt{34}-1 \] \[ -3.415<n<2.415 \]
Step 4: Count integral values of \(n\) The integers satisfying the inequality are: \[ n=-3,-2,-1,0,1,2 \] Total number of integral values: \[ 6 \]